Q. $\sum_{n = 1}^{\infty}\frac{1}{(n + 1)(n + 2)\ldots(n + k)}$ is equal to

A) $\frac{1}{(k - 1)(k - 1)!}$
B) $\frac{1}{kk!}$
C) $\frac{1}{(k - 1)k!}$
D) $\frac{1}{k!}\$
Explanation

(c)
Let $a_{n} = \frac{1}{(n + 1)(n + 2)\ldots(n + k)}$
$\Rightarrow a_{n} = \frac{1}{(k - 1)}\left( \frac{(n + k) - (n + 1)}{(n + 1)(n + 2)\ldots(n + k)} \right)$
$\Rightarrow \ \frac{1}{(k - 1)}\left( \frac{1}{(n + 1)(n + 2)\ldots(n + k - 1)} \right) - \left( \frac{1}{(n + 2)(n + 3)\ldots(n + k)} \right)$
$\therefore\ S_{n} = a_{1} + a_{2} + ... + a_{n}$
$= \frac{1}{(k - 1)}\left( \frac{1}{2 \bullet 3 \bullet 4\ldots k} - \frac{1}{(n + 2)(n + 3)\ldots(n + k)} \right)$
$\Rightarrow \ \ \lim_{n \rightarrow \infty}{S_{n} =}\frac{1}{(k - 1)k!}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.